Question

Find the inverse laplace transform of: (2 s + 4) / (s - 3)^3

Answer 1

Answer:

$e^{3t}(2t+5t^{2})$

Step-by-step explanation:

$L^{-1}[\frac{2s+4}{(s-3)^{3}} ]=$

Using the Translation theorem to transform the s-3 to s, that means multiplying by and change s to s+3

Translation theorem:$L^{1} [F(s-a)=L^{-1}[F(s)|_{s \to s-a}\\ L^{1} [F(s-a)=e^{at} f(t)$

$L^{-1}[\frac{2s+4}{(s-3)^{3}} ]=e^{3t} L^{-1}[\frac{2(s+3)+4}{s^{3}} ]$

Separate the fraction in a sum:

$e^{3t} L^{-1}[\frac{2s+10}{s^{3}} ]=e^{3t} L^{-1}[\frac{2s}{s^{3}}+\frac{10}{s^{3}} ]=e^{3t} (L^{-1}[\frac{2}{s^{2}}]+ L^{-1}[\frac{10}{s^{3}}])$

The formula for this is:

$L^{-1}[\frac{n!}{s^{n+1}} ]=t^{n}$

Modify the expression to match the formula.

$e^{3t} (2L^{-1}[\frac{1}{s^{1+1}}]+ \frac{10}{2} L^{-1}[\frac{2}{s^{2+1}}])=e^{3t} (2L^{-1}[\frac{1}{s^{1+1}}]+ 5 L^{-1}[\frac{2}{s^{2+1}}])$

Solve

$e^{3t} (2L^{-1}[\frac{1}{s^{1+1}}]+ 5 L^{-1}[\frac{2}{s^{2+1}}])=e^{3t}(2t+5t^{2} )$